Norbert W. answered • 10/17/16
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Let P be a point in the plane. Then P has coordinates (x, 18-6x-2z, z).
The vector u = 6i + j + 2k is perpendicular to the plane. The point in the plane that is closest to the origin would be OP and would be represented as OP = xi + (18-6x-2z)j + zk. This point would also be perpendicular to the plane. This means that it is a multiple of u. From this OP = s u and
x = 6s and 18 - 6x - 2z = s and z = 2s, and P(6s, 18 - 40s, 2s)
From the value for 18 - 6x - 2z = 18 -36s - 4s = 18 - 40s = s indicates that s = 18/41
Then the point P is (108/41, 18/41, 36/41).
Then the distance from the origin to the plane is √((108/41)2 + (18/41)2 + (36/41)2) = 18/√(41)
Norbert W.
In three dimensions from a point (x, y, z), the distance from the origin to that point is given by
√(x2 + y2 + z2). In this case, this is the distance from the origin to the point P in the plane, which is the same as the distance from origin to the plane, i.e. the shortest distance from the origin to the plane which is the length of the perpendicular from the origin to the plane.
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10/17/16
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10/17/16